The p-Block Elements Test

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The p-Block Elements Test - 50

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Weekly Quiz Competition

Question 1
1 / -0

The structural unit present in pyrosillicates is:

A
$$Si_3O_9^{6-}$$

B
$$SiO_4^{4-}$$

C
$$Si_2O_7^{6-}$$

D
$$(Si_2O_5^{2-})_n$$

Solution

Pyrosilicates are silicates which contain $$Si_2O_7^{6-}$$ . They are formed by joining two tetrahedral $$Si_2O_4^{4-}$$ . When two $$Si_2O_4^{4-}$$ are joined, there is removal of one oxygen atom and the two units join at the corner oxygen atom. Example is thortveitite, $$Sc_2Si_2 O_7$$ . The structure of pyrosilicate is:
Question 2
1 / -0

When $${ SO }_{ 2 }$$ is passed through acidified $${ K }_{ 2 }{ Cr }_{ 2 }{ O }_{ 7 }$$ solution:

A
the solution truns blue.

B
$${ SO }_{ 2 }$$ is reduced.

C
green $${ Cr }_{ 2 }({ SO }_{ 4 }{ ) }_{ 3 }$$ is formed.

D
the solution is decolourised.

Solution

Green $$Cr_2(SO_4)_3$$ is formed When $$SO_2$$ is bubbled through acidified solution of $$K_2Cr_2O_7$$, the orange colour of $$K_2Cr_2O_7$$ solution turns to clean green due to formation of chromium sulphate.
Hence option C is correct answer.
Question 3
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$$HCl$$ cn be oxidised by:

A
$$KMnO_4$$

B
$$Cl_2$$

C
Conc. $$H_2SO_4$$

D
$$KMnO_4$$ and Conc. $$H_2SO_4$$

Solution

$$H^+$$ in $$HCl$$ is the reactant reduced, and $$HCl$$ is the oxidizing agent. $$Zn$$ is the reactant oxidized and the reducing agent. $$Cu^{2+}$$ in $$CuCl_2$$ is the reactant reduced, and $$CuCl_2$$ is the oxidizing agent.$$Cl_2$$ is the reactant reduced and the oxidizing agent.
Question 4
1 / -0

Select the correct statements:

A
First IP of Mg is less than that of Al

B
$$O^+$$ has lower electronegativity than that of $$O^-$$

C
Electron affinity of Cl is lesser than that of F

D
Second IP of oxygen is greater than that of N

Solution

solution:
The second ionisation energy is the energy required to remove the electron from the corresponding mono-valent cation of the respective atom.

Oxygen has greater second ionisation energy than fluorine.

This is because oxygen atom has stable electronic configuration, $$2s^2, 2p^3$$ after removing one electron, the O+ shows greater ionisation energy than $$F^+$$
hence the correct option: D
Question 5
1 / -0

Blue vitriol has:

A
Ionic bond

B
Coordinate bond

C
Hydrogen bond

D
All the above

Solution

Option D is the correct answer.
$$\text{Blue vitriol$$ or $$copper sulphate pentahydrate}$$ has a molecular formula of $$CuSO_4.5H_2O.$$ of the five water molecules associated with this structure, only four are directly replaceable by $$NH_3$$, which suggest that water molecule is somehow differently attached.

Blue vitriol is $$CuSO_4.5H_2O$$ and it has all types of bonds.
The structure of $$CuSO_4.5H_2O$$ in the solid-state is shown above.
The $$Cu^{2+}$$ ions are attracted towards $$SO_4^{2−}$$ ions not only by ionic interactions (electrovalent) but also by coordinate covalent bonds.
The $$Cu^{2+}$$ ions form coordinate covalent bonds with water as well as sulfate ions.
There are covalent bonds in water and sulfate ions.
Question 6
1 / -0

The process requiring absorption of energy is :

A
$$N\rightarrow { N }^{ - }$$

B
$$F\rightarrow { F }^{ - }$$

C
$$Cl\rightarrow { Cl }^{ - }$$

D
$$H\rightarrow { H }^{ - }$$

Solution

Actually, all the elements can easily gain an electron to for ions like in H, F, Cl or N-atoms.
But, all the rest atoms except Nitrogen accepts only one electron so, they release energy and the value of electron gain enthalpy is negative.
Question 7
1 / -0

The correct order of melting point of 14th group element is:

A
$$C > Si > Ge > Pb > Sn$$

B
$$C < Si < Ge < Sn < Pb$$

C
$$Sn > Pb > Ge > Si >C$$

D
$$Si > C > Ge > Sn > Pb$$

Solution

The melting point of Group-14 decreases down the group with the exception of $$Pb$$ whose melting point is slightly higher than that of $$Sn$$.

Hence the correct order is : $$ C > Si > Ge > Sn >Pb$$
Question 8
1 / -0

Maximum covalency shown by sulphur is:

A
$$2$$

B
$$4$$

C
$$6$$

D
none

Solution

$$S=1s^22s^22P^63s^23p^4=6$$
Sulpher is the member of $$16$$ group element so it's maximum covalency is $$6$$ also in it's excited state it has $$6$$ unpaired electron thus it has $$6$$ maximum covalency for example in $$SF_{6}$$ the maximum valency is $$6$$.
Question 9
1 / -0

Which of the following is correct w.r.t. to the radii related to lead?

A
$$Pb^{4+} > Pb^{2+} > Pb$$

B
$$Pb > Pb^{2+} > Pb^{4+}$$

C
$$ Pb^{2+} > Pb > Pb^{4+}$$

D
Depends upon conditions

Solution
Question 10
1 / -0

In the following which has highest boiling point:

A
$$H I$$

B
$$H F$$

C
$$H Br$$

D
$$HCl$$

Solution

The order of boiling is $$HF>HI>HBr>HCl$$
Due to strong hydrogen bonding in HF. The boiling point of HF is very high. So, HF has highest boiling point.

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 50

Result

The p-Block Elements Test - 50

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SHARING IS CARING

Question 1

$$Si_3O_9^{6-}$$

$$SiO_4^{4-}$$

$$Si_2O_7^{6-}$$

$$(Si_2O_5^{2-})_n$$

Solution

Question 2

the solution truns blue.

$${ SO }_{ 2 }$$ is reduced.

green $${ Cr }_{ 2 }({ SO }_{ 4 }{ ) }_{ 3 }$$ is formed.

the solution is decolourised.

Solution

Question 3

$$KMnO_4$$

$$Cl_2$$

Conc. $$H_2SO_4$$

$$KMnO_4$$ and Conc. $$H_2SO_4$$

Solution

Question 4

First IP of Mg is less than that of Al

$$O^+$$ has lower electronegativity than that of $$O^-$$

Electron affinity of Cl is lesser than that of F

Second IP of oxygen is greater than that of N

Solution

Question 5

Ionic bond

Coordinate bond

Hydrogen bond

All the above

Solution

Question 6

$$N\rightarrow { N }^{ - }$$

$$F\rightarrow { F }^{ - }$$

$$Cl\rightarrow { Cl }^{ - }$$

$$H\rightarrow { H }^{ - }$$

Solution

Question 7

$$C > Si > Ge > Pb > Sn$$

$$C < Si < Ge < Sn < Pb$$

$$Sn > Pb > Ge > Si >C$$

$$Si > C > Ge > Sn > Pb$$

Solution

Question 8

$$2$$

$$4$$

$$6$$

none

Solution

Question 9

$$Pb^{4+} > Pb^{2+} > Pb$$

$$Pb > Pb^{2+} > Pb^{4+}$$

$$ Pb^{2+} > Pb > Pb^{4+}$$

Depends upon conditions

Solution

Question 10

$$H I$$

$$H F$$

$$H Br$$

$$HCl$$

Solution

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